What is the probability for each number in a set of random numbers to be the highest number in the set? I have viewed/researched many problems on this website but this is my first question.  I'm a little apprehensive by the length of the question but I have tried to explain it fully including a little of my thought process.  The problem is ...
I have n numbers in a set (N1, N2, N3 ... Nn). For each number I choose a random number (R1, R2, R3 ... Rn) between 1 and the number (e.g. 0 < R1 < N1, 0 < R2 < N2, 0 < R3 < N3 ... 0 < Rn < Nn ).  My quest is to seek the probability for each random number that it should be the largest of the random numbers in the set i.e. P(R1=max(R1 ... Rn)), P(R2=max(R1 ... Rn)) etc. up to P(Rn=max(R1 ... Rn)).
For example and by way of a Monte Carlo simulation (MCS) with a count of 10,000 I get the following ...
N1=300, N2=250, N3=450
R1 is Max(R1,R2,R3) 2576 times i.e. P(R1=Max(R1,R2,R3)) = 0.2576
R2 is Max(R1,R2,R3) 1526 times i.e. P(R2=Max(R1,R2,R3)) = 0.1526
R3 is Max(R1,R2,R3) 5898 times i.e. P(R3=Max(R1,R2,R3)) = 0.5898
Obviously the MCS will produce a slightly different result each time it is run. On 10 runs the MCS returned results of ...
  N1   N2  N3

Min 2504 1499 5854
Ave 2546 1534 5921
Max 2591 1564 5972
Interestingly using some Mathematica code as provided by the Wolfram forum the probabilities were calculated as ...
P(R1=Max(R1,R2,R3))= 0.256173
P(R2=Max(R1,R2,R3))= 0.154321
P(R3=Max(R1,R2,R3))= 0.589506
So it seems there is probably a formula to return the expected result from an infinite number of MCS runs but it's proving difficult to find it!
A helpful Wolfram forumite from another time zone promised to help me after a good night's sleep but unfortunately they deleted my thread before he woke up because I was not seeking Mathematica support!  :(
Anyway if someone could help with a formula or point to where I may find one I would be very grateful.  I need to be able to resolve this using Excel, SQL or VBA because the result needs to link with an existing SQL database so the MCS option is not really suitable.  That was only for testing and is not a viable option because I have c.150,000 number sets with 75% of them having between 5 and 12 numbers to each set and the thought of performing that number of MCS's throws my computer into a cold sweat! :)
Well if you have got this far thank you for reading my question ... it is long but hopefully most of the content is for explanation and not question.
 A: Assume we are interested in the probability that the number $R_m$ is the largest one.
The probability of this is:
$$
P(R_m)=\frac1{N_m}\sum_{i=1}^{N_m}\prod_{k\ne m}\frac{\min(i,N_k)}{N_k}.\tag1$$
Particularly for your example one obtais:
$$P_1=0.257285,\ P_2=0.590619,\ P_3=0.155248.$$
This differs slightly from your result and besides does not sum to $1$. The reason for this is the fact that the result of a trial is ambiguous if there are two or more largerst numbers, so that the events "$R_1$ is the largest number" and "$R_2$ is the largest number" do not exclude each other if $R_1=R_2$.
Actually it is very strange that you did not report about such cases, because the probability of an event with multiple "winners" in your example is $0.0031$ and one would expect about 31 such events in 10000 trials. In your statistics however they are not present. I suspect you either discard such trials or somehow assign the largest property to one of the "winners".

UPDATE for real random numbers.
In the case of real random numbers the equation (1) turns into:
$$
P(R_m)=\frac1{N_m}\int_0^{N_m}\prod_{k\ne m}\frac{\min(x,N_k)}{N_k}dx\tag2
$$
This expression can be simplified as follows. Let
$N_1\le N_2\le \dots\le N_n$. Then
$$\begin{align}
P(R_m)&=\frac1{\prod\limits_{k}N_k}
\Big[\int_0^{N_1}x^{n-1}dx+N_1\int_{N_1}^{N_2} x^{n-2}dx\\
&+N_1N_2\int_{N_2}^{N_3} x^{n-3}dx+\cdots+
N_1N_2\dots N_{m-1}\int_{N_{m-1}}^{N_m} x^{n-m}dx
\Big]\\
&=\frac{\prod_{k>m}\frac{N_m}{N_k}}{(n-m+1)}
-\sum_{i=1}^{m-1}\frac{\prod_{k>i}\frac{N_i}{N_k}}{(n-i+1)(n-i)}.
\tag3
\end{align}$$
Particularly in your example one obtains:
$$
P(R_2)=\frac{50}{324}\approx0.154321,\quad P(R_1)=\frac{83}{324}\approx0.256173,\quad P(R_3)=\frac{191}{324}\approx0.589506.
$$
in agreement with reported Mathematica result.
A: Let ask $p(R_1>R_2)$.
It will be the sum of  all the $R_1$ values multiply by the chance its bigger then $R_2$.
$p(R_1>R_2) = \sum_{i \in [N_1]}\frac{1}{N_1}\cdot \frac{i-1}{N_2}=\frac{1}{N_1\cdot N_2}\cdot \sum_{i\in [N_1]}i-1=\frac{1}{N_1\cdot N_2}\cdot \frac{N_1^2-N_1}{2}=\frac{N_1-1}{2\cdot N_2}$
From here, $P(R_1 > R_2,R_3,...,R_n)=\frac{N_1-1}{2\cdot N_2}\cdot \frac{N_1-1}{2\cdot N_3}\cdot ...\cdot \frac{N_1-1}{2\cdot N_n}$
